Algorithm to generate all possible permutations of

Question

Say I have a list of n elements, I know there are n! possible ways to order these elements. What is an algorithm to generate all possible orderings of this list? Example, I have list [a, b, c]. The algorithm would return [[a, b, c], [a, c, b,], [b, a, c], [b, c, a], [c, a, b], [c, b, a]].

I'm reading this here http://en.wikipedia.org/wiki/Permutation#Algorithms_to_generate_permutations

But Wikipedia has never been good at explaining. I don't understand much of it.

Answer 1

There is already plenty of good solutions here, but I would like to share how I solved this problem on my own and hope that this might be helpful for somebody who would also like to derive his own solution.

After some pondering about the problem I have come up with two following conclusions:

1. For the list L of size n there will be equal number of solutions starting with L₁, L₂ ... L_n elements of the list. Since in total there are n! permutations of the list of size n, we get n! / n = (n-1)! permutations in each group.
2. The list of 2 elements has only 2 permutations => [a,b] and [b,a].

Using these two simple ideas I have derived the following algorithm:

permute array
    if array is of size 2
       return first and second element as new array
       return second and first element as new array
    else
        for each element in array
            new subarray = array with excluded element
            return element + permute subarray

Here is how I implemented this in C#:

public IEnumerable<List<T>> Permutate<T>(List<T> input)
{
    if (input.Count == 2) // this are permutations of array of size 2
    {
        yield return new List<T>(input);
        yield return new List<T> {input[1], input[0]}; 
    }
    else
    {
        foreach(T elem in input) // going through array
        {
            var rlist = new List<T>(input); // creating subarray = array
            rlist.Remove(elem); // removing element
            foreach(List<T> retlist in Permutate(rlist))
            {
                retlist.Insert(0,elem); // inserting the element at pos 0
                yield return retlist;
            }

        }
    }
}

Answer 2

Here is a recursive solution in PHP. WhirlWind's post accurately describes the logic. It's worth mentioning that generating all permutations runs in factorial time, so it might be a good idea to use an iterative approach instead.

public function permute($sofar, $input){
  for($i=0; $i < strlen($input); $i++){
    $diff = strDiff($input,$input[$i]);
    $next = $sofar.$input[$i]; //next contains a permutation, save it
    $this->permute($next, $diff);
  }
}

The strDiff function takes two strings, s1 and s2, and returns a new string with everything in s1 without elements in s2 (duplicates matter). So, strDiff('finish','i') => 'fnish' (the second 'i' is not removed).

Answer 3

Here is an algorithm in Python that works by in place on an array:

def permute(xs, low=0):
    if low + 1 >= len(xs):
        yield xs
    else:
        for p in permute(xs, low + 1):
            yield p        
        for i in range(low + 1, len(xs)):        
            xs[low], xs[i] = xs[i], xs[low]
            for p in permute(xs, low + 1):
                yield p        
            xs[low], xs[i] = xs[i], xs[low]

for p in permute([1, 2, 3, 4]):
    print p

You can try the code out for yourself here: http://repl.it/J9v

Answer 4

If anyone wonders how to be done in permutation in javascript.

Idea/pseudocode

1. pick one element at a time
2. permute rest of the element and then add the picked element to the all of the permutation

for example. 'a'+ permute(bc). permute of bc would be bc & cb. Now add these two will give abc, acb. similarly, pick b + permute (ac) will provice bac, bca...and keep going.

now look at the code

function permutations(arr){

   var len = arr.length, 
       perms = [],
       rest,
       picked,
       restPerms,
       next;

    //for one or less item there is only one permutation 
    if (len <= 1)
        return [arr];

    for (var i=0; i<len; i++)
    {
        //copy original array to avoid changing it while picking elements
        rest = Object.create(arr);

        //splice removed element change array original array(copied array)
        //[1,2,3,4].splice(2,1) will return [3] and remaining array = [1,2,4]
        picked = rest.splice(i, 1);

        //get the permutation of the rest of the elements
        restPerms = permutations(rest);

       // Now concat like a+permute(bc) for each
       for (var j=0; j<restPerms.length; j++)
       {
           next = picked.concat(restPerms[j]);
           perms.push(next);
       }
    }

   return perms;
}

Take your time to understand this. I got this code from (pertumation in JavaScript)

Answer 5

Java version

/**
 * @param uniqueList
 * @param permutationSize
 * @param permutation
 * @param only            Only show the permutation of permutationSize,
 *                        else show all permutation of less than or equal to permutationSize.
 */
public static void my_permutationOf(List<Integer> uniqueList, int permutationSize, List<Integer> permutation, boolean only) {
    if (permutation == null) {
        assert 0 < permutationSize && permutationSize <= uniqueList.size();
        permutation = new ArrayList<>(permutationSize);
        if (!only) {
            System.out.println(Arrays.toString(permutation.toArray()));
        }
    }
    for (int i : uniqueList) {
        if (permutation.contains(i)) {
            continue;
        }
        permutation.add(i);
        if (!only) {
            System.out.println(Arrays.toString(permutation.toArray()));
        } else if (permutation.size() == permutationSize) {
            System.out.println(Arrays.toString(permutation.toArray()));
        }
        if (permutation.size() < permutationSize) {
            my_permutationOf(uniqueList, permutationSize, permutation, only);
        }
        permutation.remove(permutation.size() - 1);
    }
}

E.g.

public static void main(String[] args) throws Exception { 
    my_permutationOf(new ArrayList<Integer>() {
        {
            add(1);
            add(2);
            add(3);

        }
    }, 3, null, true);
}

output:

  [1, 2, 3]
  [1, 3, 2]
  [2, 1, 3]
  [2, 3, 1]
  [3, 1, 2]
  [3, 2, 1]

Answer 6

in PHP

$set=array('A','B','C','D');

function permutate($set) {
    $b=array();
    foreach($set as $key=>$value) {
        if(count($set)==1) {
            $b[]=$set[$key];
        }
        else {
            $subset=$set;
            unset($subset[$key]);
            $x=permutate($subset);
            foreach($x as $key1=>$value1) {
                $b[]=$value.' '.$value1;
            }
        }
    }
    return $b;
}

$x=permutate($set);
var_export($x);